In a survey of 1132 people, 760 people said they voted in a recent presidential election. Voting records show that 64% of eligible voters actually did vote. Given that 64% of eligible voters actually did voté, (a) find the probability tnat dimung randomly selected voters, at least 760 actually did vote. (b) What do the results from part (a) suggest? (a) P(X2760) = (Round to four decimal places as needed.) (b) What does the result from part (a) suggest? O A. People are being honest because the probability of P(x2 760) is less than 5% O B. People are being honest because the probability of P(x2 760) is at least 1%. O C. Some people are being less than honest because P(x2 760) is less than 5%. O D. Some people are being less than honest because P(x 2 760) is at least 1%.

MATLAB: An Introduction with Applications
6th Edition
ISBN:9781119256830
Author:Amos Gilat
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Chapter1: Starting With Matlab
Section: Chapter Questions
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**Transcription for Educational Website:**

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In a survey of 1132 people, 760 people said they voted in a recent presidential election. Voting records show that 64% of eligible voters actually did vote. Given that 64% of eligible voters actually did vote, (a) find the probability that among 1132 randomly selected voters, at least 760 actually did vote. (b) What do the results from part (a) suggest?

(a) \( P(X \geq 760) = \) [Round to four decimal places as needed.]

(b) What does the result from part (a) suggest?

- O A. People are being honest because the probability of \( P(X \geq 760) \) is less than 5%.
- O B. People are being honest because the probability of \( P(X \geq 760) \) is at least 1%.
- O C. Some people are being less than honest because \( P(X \geq 760) \) is less than 5%.
- O D. Some people are being less than honest because \( P(X \geq 760) \) is at least 1%.

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**Explanation:**

In this problem, you are asked to calculate the probability that at least 760 out of 1132 randomly selected people voted, given that 64% of eligible voters actually did vote. This requires using statistical methods to determine how likely it is to observe 760 or more votes if the true voting rate is 64%.

The analysis helps identify whether people might not be accurately reporting their voting behavior. If the calculated probability is very low (below 5%), it suggests some level of dishonesty in the survey responses. Conversely, a higher probability would indicate the responses are likely honest. 

Choose the correct option based on the calculated probability in part (a).
Transcribed Image Text:**Transcription for Educational Website:** --- In a survey of 1132 people, 760 people said they voted in a recent presidential election. Voting records show that 64% of eligible voters actually did vote. Given that 64% of eligible voters actually did vote, (a) find the probability that among 1132 randomly selected voters, at least 760 actually did vote. (b) What do the results from part (a) suggest? (a) \( P(X \geq 760) = \) [Round to four decimal places as needed.] (b) What does the result from part (a) suggest? - O A. People are being honest because the probability of \( P(X \geq 760) \) is less than 5%. - O B. People are being honest because the probability of \( P(X \geq 760) \) is at least 1%. - O C. Some people are being less than honest because \( P(X \geq 760) \) is less than 5%. - O D. Some people are being less than honest because \( P(X \geq 760) \) is at least 1%. --- **Explanation:** In this problem, you are asked to calculate the probability that at least 760 out of 1132 randomly selected people voted, given that 64% of eligible voters actually did vote. This requires using statistical methods to determine how likely it is to observe 760 or more votes if the true voting rate is 64%. The analysis helps identify whether people might not be accurately reporting their voting behavior. If the calculated probability is very low (below 5%), it suggests some level of dishonesty in the survey responses. Conversely, a higher probability would indicate the responses are likely honest. Choose the correct option based on the calculated probability in part (a).
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